From pg. 113 of Categories for the Working Mathematician:
Problem: I understand everything about this proof except the red underlined: why does
$$\tau \text{ a cone } \implies fh = gh?$$
Especially since $\prod_u F_{\text{cod} u}$ needn't even be an $F(i)$ for some $i \in \text{ob}(J)$.
A cone $\tau$ consists of a collection of morphisms indexed by the objects $i \in ob(J)$: $\tau_i:C \rightarrow Fi$. These induce a morphism $C \rightarrow \prod_{ob(J)} Fi$ by the usual universal property of a product, which we denote $h: C \rightarrow \prod_{ob(J)} Fi$.
If you understand how $f$ and $g$ are constructed as morphisms $\prod_{ob(J)} Fi \rightarrow \prod_{mor(J)} Fcod(u)$ (remember that a morphism into a product is defined by its composition with all projections) then we can see that $fh$ and $gh$ are both morphisms $C \rightarrow \prod_{ob(J)} Fi \rightarrow \prod_{mor(J)} Fcod(u)$.
To show that $fh$ and $gh$ are equal, we show that their composites with all the projections from $\prod_{mor(J)} Fcod(u)$ are equal. To calculate $\pi_u fh$ and $\pi_u gh$, first use the definitions of $f$ and $g$ and then use the fact that the components of $h$ were the maps of a cone.